Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
给链表添加一个临时的头结点, 这样操作更方便。其实大部分链表问题,添加一个头结点，都会简化后面的操作 本文地址
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/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class
Solution {
public
:
ListNode *swapPairs(ListNode *head) {
ListNode tmphead(0); tmphead.next = head;
ListNode *pre = &tmphead, *p = head;
while
(p && p->next)
//p 和 p->next是待交换的两个节点，pre是p的前一个节点
{
pre->next = p->next;
p->next = p->next->next;
pre->next->next = p;
pre = p;
p = p->next;
}
return
tmphead.next;
}
};
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本文转自tenos博客园博客，原文链接：http://www.cnblogs.com/TenosDoIt/p/3793641.html，如需转载请自行联系原作者
